Guaranteeing negative result when left shifting a negative number in two's complement?

Question

Assuming a negative binary number is represented in two's complement, how can we guarantee the sign is preserved?

Let's say we represent a decimal number -5 in four bits: 1011, and want to left-shift one place to multiply by 2:

1011 << 1

This operation returns 0110, which is 6, not -10 as we would have hoped.

(I assume this is only the case for negative numbers whose second bit is 0, ie negative numbers close to the smallest representable negative number of some range)

Answer 1

OP here. I found the answer to my question.

Shifting left may trigger arithmetic overflow.

The range of numbers expressible by a two's complement system is from -(2^(n-1)) to 2^(n-1)-1, where n is the number of bits available, including the sign bit (MSB). So, in the above example where each number uses 4 bits, the range of possible values is -8 to 7, inclusive.

Shifting left by m bits will multiply the number by 2^m. So, in the above example -5 << 1 would yield -10, which is beyond the range of possible numbers in a 4-bit signed representation – that's an overflow.

1111 << 1 == 1110 // -1 * 2 is -2
1110 << 1 == 1100 // -2 * 2 is -4
1101 << 1 == 1010 // -3 * 2 is -6
1100 << 1 == 1000 // -4 * 2 is -8
1011 << 1 == 0110 // overflow
1010 << 1 == 0100 // overflow
1001 << 1 == 0010 // overflow
1000 << 1 == 0000 // overflow

In conclusion, while using ASL to multiply by powers of two, it is important to make sure the product lies within the range of possible values.